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Date November 2016 Marks available 3 Reference code 16N.2.SL.TZ0.T_5
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Calculate Question number T_5 Adapted from N/A

Question

A farmer owns a plot of land in the shape of a quadrilateral ABCD.

AB=105 m, BC=95 m, CD=40 m, DA=70 m and angle DCB=90.

N16/5/MATSD/SP2/ENG/TZ0/05

The farmer wants to divide the land into two equal areas. He builds a fence in a straight line from point B to point P on AD, so that the area of PAB is equal to the area of PBCD.

Calculate

the length of BD;

[2]
a.

the size of angle DAB;

[3]
b.

the area of triangle ABD;

[3]
c.

the area of quadrilateral ABCD;

[2]
d.

the length of AP;

[3]
e.

the length of the fence, BP.

[3]
f.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

(BD=) 952+402    (M1)

 

Note:     Award (M1) for correct substitution into Pythagoras’ theorem.

 

=103 (m) (103.077, 2517)    (A1)(G2)

[2 marks]

a.

cosBˆAD=1052+702(103.077)22×105×70     (M1)(A1)(ft)

 

Note:     Award (M1) for substitution into cosine rule, (A1)(ft) for their correct substitutions. Follow through from part (a).

 

(BˆAD)=68.9 (68.8663)    (A1)(ft)(G2)

 

Note:     If their 103 used, the answer is 68.7995

 

[3 marks]

b.

(Area of ABD=)12×105×70×sin(68.8663)    (M1)(A1)(ft)

 

Notes:     Award (M1) for substitution into the trig form of the area of a triangle formula.

Award (A1)(ft) for their correct substitutions.

Follow through from part (b).

If 68.8° is used the area =3426.28 m2.

 

=3430 m2 (3427.82)    (A1)(ft)(G2)

[3 marks]

c.

area of ABCD=12×40×95+3427.82    (M1)

 

Note:     Award (M1) for correctly substituted area of triangle formula added to their answer to part (c).

 

=5330 m2 (5327.83)    (A1)(ft)(G2)

[2 marks]

d.

12×105×AP×sin(68.8663)=0.5×5327.82    (M1)(M1)

 

Notes:     Award (M1) for the correct substitution into triangle formula.

Award (M1) for equating their triangle area to half their part (d).

 

(AP=) 54.4 (m) (54.4000)    (A1)(ft)(G2)

 

Notes:     Follow through from parts (b) and (d).

 

[3 marks]

e.

BP2=1052+(54.4000)22×105×(54.4000)×cos(68.8663)    (M1)(A1)(ft)

 

Notes:     Award (M1) for substituted cosine rule formula.

Award (A1)(ft) for their correct substitutions. Accept the exact fraction 53147 in place of cos(68.8663).

Follow through from parts (b) and (e).

 

(BP=) 99.3 (m) (99.3252)    (A1)(ft)(G2)

 

Notes:     If 54.4 and cos(68.9) are used the answer is 99.3567

 

[3 marks]

f.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.2—2d and 3d trig, sine rule, cosine rule, area
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Topic 3— Geometry and trigonometry

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